I'll begin by stating a few assumptions. I'm going to assuming that the rain is falling straight down, at a constant rate. I'm also going to assume that if we are standing still, only our head and shoulders get wet, not our front or back. With those in place, lets start by formulating the expression for how wet we would get if we stood still. Well, take

\[\Delta W_{top} - \text{the change in water (in liters) on a person} \]

\[\rho - \text{the density of water in the air in liters per cubic meter}\]

\[A_t - \text{top area of a person}\]

\[\Delta t - \text{time elapsed}\]

Intuition suggests that the rate at which raindrops hit our top, times the area of our top, times the time we stand in the rain, will give us the change in water. In an equation,

\[ \Delta W_{top} = \rho A_t v_r \Delta t \]

Note that whatever expression we generate for how wet we get when moving will have to reduce to this form in the limit that we're not moving. This will be a good check for us. Next, we need to define a few additional measures:

\[d - \text{distance we have to travel in the rain}\]

\[v_r - \text{raindrop velocity}\]

\[A_f - \text{front area of a person}\]

\[W_{tot} - \text{total amount of water in liters we get hit with} \]

Well, no mater how fast we run the rain will keep hitting us on the top of our heads, so we're going to have our standing still term, plus another term for how much hits us when running. How do we consider that? Well, when we run, we're cutting into the swath of rainy air in front of ourselves. We'll get hit on our frontside by all the additional raindrops in that stretch we carve out. Mathematically, if we travel some distance delta x in a time delta t, we'll get hit with an additional amount of water

\[ \Delta W = \rho A_f \Delta x \]

\[ \Delta W = \rho A_f v \Delta t \]

We combine our two terms to get

\[\Delta W_{tot} = \rho \Delta t (A_t v_r + A_f v)\]

Note that if we stop walking (v goes to zero) we'll return to our stationary expression. Next I'll take the delta t over to the other side, turn that into a derivative, and integrate to get the total water hitting us, not just the change for some delta t. Of course, since everything else in the equation is constant, this is the equivalent of dropping the deltas,

\[W_{tot} = \rho (A_t v_r + A_f v) t \]

\[W_{tot} = \rho (A_t v_r + A_f v)\frac{d}{v}\]

\[W_{tot}= \rho d (A_t \frac{v_r}{v} + A_f)\]

where I substituted t = d/v, and did some simplification. Now, lets look at the qualitative features of this result. First, we have two terms, a constant and a term that depends inversely on the velocity of motion. This means that the faster we go, the less wet we get (I'll plot this in a bit), but also that there's a threshold wetness you cannot avoid. This threshold represents the amount of rain in a human sized channel between where you start and where you end. Also note that as velocity goes to zero, i.e. we stop moving, how wet we get goes to infinity. That is, if we're going to stand in the rain forever we're going to keep getting wet. What is the term in 1/v? It's the amount of rain that hits you on the top of your head! So what we've derived is that for a fixed distance how wet you get on the front is fixed, and by moving faster you can make less rain hit you on the top of your head.

Now, lets figure out what some reasonable numbers are, and plot this function. A few months back when discussion human radiation I estimated my area as a cylinder with a height of 1.8 m and a radius of .14 m. This gives a top area of A_t = .06 m^2 and a front cross section area (note, this is not the cylinder area, but my cross section that will be exposed to the rain!) of .5 m^2. As for raindrop velocity, well in my first post on this blog I calculated the terminal velocity of what I described as a medium sized raindrop as 6 m/s, and since water drops reach that while going over niagara falls, we can assume that our raindrops are falling at terminal velocity.

Finally, I need to estimate the water density. In a medium-to-heavy rain I would say it takes about 45 s to get a sidewalk square totally wet. Let's assume a raindrop wets an area of sidewalk equal to twice the cross section of the raindrop. I used a raindrop of 1.5mm radius, so that's 7*10^-6 m^2 cross section. Now, a sidewalk square is about 2/3 m x 2/3m (about 2 ft x 2ft), so we need ~32000 drops! The volume of a 1.5mm drop is 1.4^-8 m^3, so we have a volume of 4.5*10^-4 m^3 = .45 liters. Now, take our stationary expression from above. This allows us to solve for \rho. Set delta W equal to .45 liters and substitute the rest of the numbers we've generated.

\[ \frac{\Delta W_{top}}{A_t v_r \Delta t}= \rho \]

\[ \rho = \frac{.45 liters}{(.44 m^2)( 6 m/s )(45 s)} = .004 liters/m^3 \]

We've found our water density, .004 liters/m^3. Having done this, we can plug numbers into our final equation above and find

\[W_{tot}= d (\frac{(.001 liters/s)}{v} + .002 liters/m)\]

This scales linearly with distance, so lets pick something reasonable, say 100m, and if you want the result for another other distance just scale the results appropriately. Thus

\[W_{tot}= (\frac{(.1 liters\text{*}m/s)}{v} + .2 liters)\]

Finally we can plot this.

Plot of how wet you get vs. how fast you run. The blue line is the actual curve and the red line is the theoretical least wet asymptote. |

I've chosen .5 m/s (~1mph, a meander) and 11 m/s (slightly faster than the world record for the 100m dash) as my starting and ending points on the velocity. The blue line is the curve I calculated, and the red line represents the theoretical minimum, the 'wetness threshold' if you will. So you see that if you are Usain Bolt, you can reduce how wet you get by almost a factor of two by going from a meander to a sprint!

Now, there's more I could say about this (what if the rain isn't coming straight down? What is my best speed if I have an umbrella?), but I think that's enough for tonight. I've come out with a theoretically satisfying result that concurs with experiment. Anytime that happens that's a good day for the theorists.

Interesting! It also explains why the experiments have produced so varying results. Consider the change in pace from a brisk walk to a jog, say from 2 to 4 m/s. That would mean a change in water impact from .25 to .225 litres. This change in wetness, a mere 10%, is most likely within the error bars of most of the experiments.

ReplyDeleteThat's a good point that I hadn't thought about! Though, to be fair, if we let the raindrop velocity increase, the difference will increase. However, given the kinds of experiments we're talking about, that's probably still not enough to be outside the error bars.

ReplyDeleteOf course, a human isn't a cylinder that's moving in smooth translations. A_t and (to a lesser extent) A_f vary with time as a person moves and swings their arms and legs back and forth. You could probably simplify this and only use the average A_t and A_f, but those average figures will be dependent on how the person is moving. E.g. if someone is running through the rain, they extend their arms and legs much further than someone who's walking, and thus they have a greater A_t.

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